Answer
$x=2.53$
Work Step by Step
We test $f(x)=x^{3}+x^{2}+x-4$
on nonnegative integer values of x, and find that
$f(2) $is negative, $f(3)$ is positive $\Rightarrow$ by the IVT, a real zero is inside $[2,3].$
Subdividing $[2,3]$ into 10 subintervals, calculating f( endpoints), we find
$f(2.5) $is negative, $f(2.6)$ is positive $\Rightarrow$ by the IVT, a real zero is inside $[2.5,2.6].$
Subdividing again, we find
$f(2.53) $is negative, $f(2.54)$ is positive $\Rightarrow$ by the IVT, a real zero is inside $[2.53,2.54].$
The real zero is $x=2.53...$
To two decimal places, $x=2.53$
